f5 maths

It touches x-axis, so when sub y=0 (x-axis is y=0), it has a double root.

Choice a)
x^2-6x+4=0 does not has a double root

Choice b)
x^2-6x-4=0 does not has a double root

Choice c)
x^2-6x+9=0 does has a double root of x=3

Choice d)
x^2-6x-9=0 does not has a double root


2007-02-26 20:50:50 補充：
The answer is C.

因為與x軸相切,只要將各式化為圓的標準方程(即(x-a)^2+(y-b)^2=r^2),
看看半徑是否等於圓心縱坐標的絕對值便可判斷
(即r>b的絕對值,與x軸相交;r=b的絕對值,與x軸相切;r



2,所以與x軸相交

b. x^2 +y^2-6x+4y-4=0
(x-3)^2+(y+2)^2=17
因為17的平方根>2,所以與x軸相交

c. x^2 +y^2-6x+4y+9=0
(x-3)^2+(y+2)^2=2^2
因為2=2,所以與x軸相切

d. x^2 +y^2-6x+4y-9=0
(x-3)^2+(y+2)^2=22
因為22的平方根>2,所以與x軸相交

即答案是c

2007-02-27 04:57:46 補充：
<<不知為何小了一部分>>(即r大於b的絕對值,與x軸相交;r=b的絕對值,與x軸相切;r小於b的絕對值,與x軸沒有交點)a. x^2 +y^2-6x+4y+4=0(x-3)^2+(y+2)^2=3^2因為3>2,所以與x軸相交